Basically, they show that you can use SAT to solve Sudoku, and then claim that this makes Sudoku NP-complete. (All it shows is that Sudoku is in NP.)
People make similar mistakes often when they want to show that a certain problem isn't solvable in linear time, and they try to show that sorting can solve your problem. But it's the wrong direction.
Wait, did you mess up the direction here too, or am I confused? If you reduce problem A to B, then it means B is at least as hard as A, because solving it would solve A. Which certainly means in this case that Sudoku is NP-hard. And it doesn't (without introducing additional facts) imply Sudoku is in NP either. I don't see anything wrong here, do you?
That is, you've shown that an oracle that solves any SAT problem in constant time can solve any Sudoku in polynomial time.
The more difficult side is to show that for any SAT instance, you can reduce it to a Sudoku.
Really proving that you can use SAT to solve Sudoku is not a great or interesting result; since Sudoku is a decision problem it is very clear that it is in NP. Or see that verifying that a Sudoku solution is correct is achievable in polynomial time.
> Wait, did you mess up the direction here too, or am I confused? If you reduce problem A to B, then it means B is at least as hard as A, because solving it would solve A.
Using SAT to solve Sudoku is a reduction of Sudoku to SAT. The order of the problems names switches depending on how you write it.
Reducing unknown-complexity to NP-complete means you can bound the complexity by above, but not by below. I can reduce binary integer addition to SAT, which means that binary integer addition is no harder than SAT... but we also know by other algorithms that it is in fact easier than SAT.
To bound by below, you have to reduce NP-complete (or NP-hard suffices) to unknown-complexity.
Lololol